Question

Physical activity of obese young adults. In a study on the physical activity of young adults, pediatric researchers measured overall physical activity as the total number of registered movements (counts) over a period of time and then computed the number of counts per minute (cpm) for each subject (International Journal of Obesity, Jan. 2007). The study revealed that the overall physical activity of obese young adults has a mean of μ = 320 cpm μ = 320   cpm and a standard deviation of σ = 100 c p m . σ = 100 c p m . (In comparison, the mean for young adults of normal weight is 540 cpm.) In a random sample of n = 100 n = 100 obese young adults, consider the sample mean counts per minute, ¯ x x ‾ . Describe the sampling distribution of ¯ x x ‾ . What is the probability that the mean overall physical activity level of the sample is between 300 and 310 cpm? What is the probability that the mean overall physical activity level of the sample is greater than 360 cpm?

Answer 1

Solution:

Given: The overall physical activity of obese young adults has a mean of μ = 320 cpm and a standard deviation of σ = 100 c p m .

Sample size = n = 100

Part a) Describe the sampling distribution of sample mean .

Since sample size = n = 100 is large, the sampling distribution of sample mean is approximately Normal with mean of sample mean is: with a standard deviation of sample mean is:

Part b) What is the probability that the mean overall physical activity level of the sample is between 300 and 310 cpm?

That is:

Find z scores:

Thus we get:

Look in z table for z = -2.0 and 0.00 as well as for z = -1.0 and 0.00 and find area.

Thus from table , we get:
P( Z < -2.00) = 0.0228

and P( Z < -1.00) =0.1587

Thus

Part c) What is the probability that the mean overall physical activity level of the sample is greater than 360 cpm?

Find z score:

Thus we get:

From z table, as P( Z < 3.49) = 0.9998 , P(Z < 4.00) would be approximately equal to 1.0000, that is: P( Z< 4.00) = 1.0000

Thus